Find the equation of the hyperbola satisfying the give conditions: Foci $(\pm 4,\,0),$ the latus rectum is of length $12$

Foci $(\pm 4,\,0),$ the latus rectum is of length $12$

Here, the foci are on the $x-$ axis.

Therefore, the equation of the hyperbola is of the form $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$

since the foci are $(\pm 4,\,0)$,  $c=4$

Length of latus rectum $=12$

$\Rightarrow \frac{2 b^{2}}{a}=12$

$\Rightarrow b^{2}=6 a$

We know that $a^{2}+b^{2}=c^{2}$

$\therefore a^{2}+6 a=16$

$\Rightarrow a^{2}+6 a-16=0$

$\Rightarrow a^{2}+8 a-2 a-16=0$

$\Rightarrow(a+8)(a-2)=0$

$\Rightarrow a=-8,2$

since a is non-negative, $a=2$

$\therefore b^{2}=6 a=6 \times 2=12$

Thus, the equation of the hyperbola is $\frac{x^{2}}{4}-\frac{y^{2}}{12}=1$